2020-I-12

By the property of similar triangles, we have

$\begin{array}{rcl}{\displaystyle \frac{r_1}{15}}& =& {\displaystyle \frac{12}{36}}\\ r_1& =& 5\end{array}$

Also,

$\begin{array}{rcl}{\displaystyle \frac{r_2}{15}}& =& {\displaystyle \frac{24}{36}}\\ r_2& =& 10\end{array}$

Therefore, the volume of the middle part

$\begin{array}{cl}=& {\displaystyle \frac{1}{3}}\pi (10{)}^2(24)–{\displaystyle \frac{1}{3}}\pi (5{)}^2(12)\\ =& 700\pi {\text{ cm}}^3\end{array}$

Consider the slant height of the upper part.

$\begin{array}{rcl}{\ell }_1^2& =& 5^2+{12}^2\\ {\ell }_1^2& =& 169\\ {\ell }_1& =& 13\text{ cm}\end{array}$

Consider the slant height of of the middle part.

$\begin{array}{rcl}({\ell }_2+13{)}^2& =& {10}^2+{24}^2\\ ({\ell }_2+13{)}^2& =& 676\\ {\ell }_2+13& =& 26\\ {\ell }_2& =& 13\end{array}$

Therefore, the curved surface area of the middle part

$\begin{array}{cl}=& \pi (10)(26)–\pi (5)(13)\\ =& 195\pi {\text{ cm}}^2\end{array}$